Is the set $A = \left\{ x \in \mathbb{R}\, |\, x^2 \text{ is rational}\right\}$ countable?

Question

Is the set $A = \left\{ x \in \mathbb{R} \,|\, x^2 \text{ is rational}\right\}$ countable?

I know that the set of all rational numbers is countable. But for some irrational numbers, $x^2$ is rational.

Example: $(\sqrt2)^2$ is rational.

So that the set $A$ includes all rational numbers and some irrational numbers.

Then how can we prove that the set $A$ is countable?

Answer 1

The range $R$ of$$\begin{array}{ccc}\{x\in\Bbb Q\mid x\geqslant0\}&\longrightarrow&\Bbb R\\x&\mapsto&\sqrt x\end{array}$$is a countable set (since its domain is countable) and $A=R\cup(-R)$. Therefore, $A$ is countable.

Answer 2

Yes, because these squares are all positive rational numbers and each positive rational number has exactly two square roots. This means that for every rational number $q>0$, there are two real numbers whose square is $q$. (Zero is the exception, with only one square root).

For example, the rational number 2 has two square roots $\pm \sqrt{2}$. The rational number $\frac{7}{3}$ has two square roots $\pm \sqrt{7/3}$, the rational number 16 has two square roots $\pm 4$, and so on.

If you tally up all of these square roots, you get the set $A$ of real numbers whose square is any rational number. There are two roots for every rational number, and there are countably many rational numbers, so this set $A$ is countable.